Team:USP-UNESP-Brazil/Plasmid Plug n Play/Modeling

From 2012.igem.org

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<ul>
<ul>
<li><span class="math">$k_{degRNA} = 1/350$</span> (1/sec).</li>
<li><span class="math">$k_{degRNA} = 1/350$</span> (1/sec).</li>
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<li><span class="math">$k_{transc} = 40$</span> (bp/sec)  for T7 promoter.</li>
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<li><span class="math">$k_{transc} = 40$</span> (bp/sec) - for T7 promoter.</li>
<li><span class="math">$k_{transl} = 15$</span> (aa/sec).</li>
<li><span class="math">$k_{transl} = 15$</span> (aa/sec).</li>
<li><span class="math">$k_{degProt} = 0.0167/60$</span> (Prot/sec) Average protein degradation.</li>
<li><span class="math">$k_{degProt} = 0.0167/60$</span> (Prot/sec) Average protein degradation.</li>
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<li><span class="math">$n_{bp} = 1032 (base pair)$</span> for CRE.</li>
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<li><span class="math">$n_{bp} = 1032 $</span> - for CRE.</li>
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<li><span class="math">$n_{aa} = n_{bp}/3 = 344(aa) $</span> for CRE.</li>
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<li><span class="math">$n_{aa} = n_{bp}/3 = 344 $</span> - for CRE.</li>
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<li><span class="math">$n_{bp} = 1119 (base pair)$</span> for FLP.</li>
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<li><span class="math">$n_{bp} = 1119 $</span> - for FLP.</li>
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<li><span class="math">$n_{aa} = n_{bp}/3 = 373 (aa) $</span> for FLP.</li>
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<li><span class="math">$n_{aa} = n_{bp}/3 = 373 $</span> - for FLP.</li>
</ul>
</ul>
</p>
</p>
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\end{align}
\end{align}
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<p>For a 800 <span class="math"><em>b</em><em>p</em></span> gene and 50 <span class="math"><em>μ</em><em>L</em></span> of solution we have: <br /><span class="math">$n_{mols} = \frac{m_{dna}}{800*6*10^{23}.10^{-12}} \simeq m_{dna} 2*10^{-15}$</span><br /> and <br /><span class="math">$[So] = \frac{m_{dna} 2*10^{-15}}{50*10^{-6}}  
+
For a 800 <span class="math"><em>b</em><em>p</em></span> gene and 50 <span class="math"><em>μ</em><em>L</em></span> of solution we have: <br /><span class="math">$n_{mols} = \frac{m_{dna}}{800*6*10^{23}.10^{-12}} \simeq m_{dna} 2*10^{-15}$</span><br /> and <br /><span class="math">$[So] = \frac{m_{dna} 2*10^{-15}}{50*10^{-6}}  
\simeq m_{dna} 0.4*10^{-10} M = 0.04 m_{dna} nM$</span><br /> This means, for example, that in order to obtain <span class="math">10</span> <span class="math"><em>n</em><em>M</em></span> of concentration <span class="math">250</span> <span class="math"><em>n</em><em>g</em></span> of DNA are needed in a solution of <span class="math">50<em>μ</em><em>L</em></span>.</p>
\simeq m_{dna} 0.4*10^{-10} M = 0.04 m_{dna} nM$</span><br /> This means, for example, that in order to obtain <span class="math">10</span> <span class="math"><em>n</em><em>M</em></span> of concentration <span class="math">250</span> <span class="math"><em>n</em><em>g</em></span> of DNA are needed in a solution of <span class="math">50<em>μ</em><em>L</em></span>.</p>

Revision as of 01:47, 26 September 2012